How Is the Electric Field Intensity Calculated on a Charged Sphere's Surface?

[BvU]

And when we flash back to post #4 ...

 

[BvU]

That's flashing back even further !

But the same link has something to say about the derivation too, I seem to remember. The name Gauss comes to mind !
I take it you can look around there all on your own (with a little patience, perhaps ...) , so -- like a good helpdesk worker -- I can now safely close this ticket, I hope ?

 

[gracy]

Do you remember Coulomb's Law?

$F$=$\frac{K Qq}{r^2}$

 

[haruspex]

I am still clueless about my question in op.
$E$=$\frac{\left(\frac{1}{4πε0}\frac{q}{R}\right)^2}{\frac{q}{4πε0}}$

They took the equation $E$=$\frac{1}{4πε_0}\frac{q}{R^2}$ and rewrote it in that form. You can check they are equivalent. This was useful because the numerator is the same as V² and the denominator only involves the charge and some constants. Since V and q are known, this gives a quick way of finding E.